Problem: What is the average rate of change of $h(x)=\dfrac{1}{10-x}$ over the interval $5\le x \le8$ ?
Explanation: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We will need to know the values of $h(5)$ and $h(8)$ to find the slope. $\begin{aligned} h(5)&=\dfrac{1}{10-5} \\\\ &=\dfrac{1}{5} \\\\\\ h(8)&=\dfrac{1}{10-8} \\\\ &=\dfrac{1}{2} \\\\\\ \dfrac{h(8)-h(5)}{8-5}&=\dfrac{\dfrac{1}{2}-\dfrac{1}{5}}{3} \\\\ &=\dfrac{\left(\dfrac{3}{10}\right)}{3} \\\\ &=\dfrac{1}{10} \end{aligned}$ The average rate of change of $h$ over the interval $5\le x \le8$ is $\dfrac{1}{10}$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{3}{5}$ $\frac{4}{5}$ $y$ $x$ $(5,h(5))$ $(8,h(8))$ secant line